_ n = 1 ^ _ k = 1 ^ n - 1 k 2^ n + k is equal to

MCQ

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

A
$\frac {2}{9}$
✓
$\frac {4}{9}$
C
$\frac {4}{3}$
D
$\frac {2}{3}$

✓

Answer

Correct option: B.

$\frac {4}{9}$

b
$\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} \left( {\frac{1}{{{2^1}}} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} + ...... + \frac{{n - 1}}{{{2^{n - 1}}}}} \right)$

$\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} \left( {2 - \frac{{n + 1}}{{{2^{n - 1}}}}} \right)$

$\sum\limits_{n = 1}^\infty {\frac{1}{{{2^{n - 1}}}} - \frac{{n + 1}}{{{2^{2n - 1}}}}} = \frac{4}{9}$

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